<p>Evaluate \(\displaystyle\int_1^4\frac{dx}{x+\sqrt{x}}\) [JEE Main 2019]</p>
Step-by-Step Solution
Key Concept: Factor: 1/(x+\sqrt{x}) = 1/(\sqrt{x}(\sqrt{x}+1)). Let t=\sqrt{x}, dt=1/(2\sqrt{x})dx \to 2dt/(t+1).
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<p>Let $t=\sqrt{x}\Rightarrow x=t^2$, $dx=2t\,dt$. Limits: $x=1\to t=1$; $x=4\to t=2$.</p>
<p>$$\int_1^2\frac{2t\,dt}{t^2+t}=\int_1^2\frac{2\,dt}{t+1}=2[\ln(t+1)]_1^2=2(\ln 3-\ln 2)=2\ln\frac{3}{2}$$</p>
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Correct Answer: A